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Let $X$ be a scheme over an algebraically closed field $k$ of positive characteristic $p$. Recall that the absolute Frobenius morphism $F : X \to X$ is the map which is the identity on points and the $p^{th}$ power morphism on functions. Recall also that we say that $X$ is Frobenius split if there is an $\mathcal O_X$-linear morphism splitting the $p^{th}$ power morphism $ \mathcal O_X \to F_* \mathcal O_X $.

Now, whenever one sees the definition of Frobenius splitting, it is always stated for an algebraically closed field $k$. However, the definitions above make perfectly good sense for any scheme over $\mathbb F_p$, and in fact many Frobenius-split schemes, eg flag varieties, are "split over $\mathbb F_p$" in the sense that the Frobenius splitting is the appropriate base-change of a morphism $F_* \mathcal O_X \to \mathcal O_X$ for a scheme $X$ over $\mathbb F_p$. (Although I defined the Frobenius morphism above for schemes over $k$ it also makes sense for any scheme over $\mathbb F_p$). My question is: Why is the definition of Frobenius splitting always stated for schemes over an algebraically closed field, when one can state it more generally for schemes over any field containing $\mathbb F_p$? Is this just convention or is there a deeper reason?

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Jantzen's book (Reps. of alg. groups), chapter F discusses Frobenius splittings over perfect fields ("we restrict...to perfect grounds fields for the sake of simplicity"). After a little while it starts to assume k alg. closed, saying "the main results in the end will usually easily extend to arbitrary k" (e.g. Frob splitting for flag varieties and Schubert varieties). –  fherzig Apr 3 '12 at 22:07

2 Answers 2

up vote 12 down vote accepted

There is of course the commutative algebra perspective on Frobenius splittings as well. Indeed, the general case there is much more general than what I think you are even considering. Let me make some comments on various generalizations of the notion of Frobenius splittings.

The perfect field case: Essentially everything works without change as David Speyer already mentions. Of course, there are the usual complications with working over non-algebraically closed field, but I am completely confident that there are no problems with the theory (many of my papers on this subject work in this generality, or the next one). One problem of course is with the differentials, if you are working at non-$k$-points, things get slightly more complicated.

The $F$-finite field case: A field $k$ is called $F$-finite if the $k$ is a finitely generated $k^p$ vector space. This automatically holds for any residue field of a variety of finite type over a perfect field (in particular, the case that David Speyer mentioned above). A variety over an $F$-finite field is still fairly well behaved. In particular, such a variety $X$ is itself $F$-finite, meaning the Frobenius morphism: $$F : X \to X$$ is a finite map. I don't think the Cartier isomorphism works without doing some relative Frobenius stuff, but, things still mostly work.

In particular, there is still a trace map: $$T : F_* \omega_X \to \omega_X$$ Here's how you see it. Fix $\omega_X$ on $X$ coming from the structural map to $k$ (in general, this is the first non-zero cohomology of $g^! k$ where $g : X \to k$ is the structural map). There is a natural map $$\mathcal{H}om_X(F^e_* O_X, \omega_X) \to \omega_X.$$ This is the evaluation at $1$ map. On the other hand, it is not difficult to show that in the context described, $\mathcal{H}om_X(F^e_* O_X, \omega_X) \cong F_* \omega_X$ (critically using the Frobenius is finite). Thus we have obtained another trace map.

Notice, this map works for all $X$, not just normal $X$ (it even works for non-reduced schemes of finite type over $k$). This map agrees with the map in the algebraically closed / perfect $k$ case, at least up to multiplication by a unit (in that isomorphism above, I have to make a choice).

Standard arguments then imply, that for normal $X$, we have that Frobenius splittings of $X$ come from sections of $H^0(X, O_X((1-p)K_X)$. Again, let me briefly describe this:

Set $U = X_{\text{reg}} \subseteq X$, the regular locus. Then we have that \begin{align} & & \mathcal{H}om_X(F_* O_X, O_X) \newline & \cong & i_* \mathcal{H}om_U(F_* O_U, O_U) \newline & \cong & i_* \mathcal{H}om_U(F_* (\omega_U^p), \omega_U) \newline & \cong & i_* \mathcal{H}om_{F_* O_U}(F_* \omega_U^p, \mathcal{H}om_U(F_* O_U, \omega_U))\newline & \cong & i_* F_* \mathcal{H}om_U(\omega_U^p, \omega_{U})) \newline & \cong & i_* F_* O_U( (1-p)K_U) \newline & \cong & F_* O_X( (1-p)K_X).\end{align}

The map $i : U \to X$ is the inclusion and the isomorphsims involving adding or removing the $i_*$ just come because the sheaves are reflexive / S2.

In particular, Frobenius splittings can still be viewed as divisors linearly equivalent to $(1-p)K_X$. The only problem is that we don't quite have the same nice local differential form picture as far as I know (at least, it hasn't been written down).

The $F$-finite scheme case: Now I want to consider schemes not of finite type over a base field, but which are still $F$-finite, in particular, $$F : X \to X$$ is a finite map. Note by a result of Kunz, such schemes are automatically locally excellent, and by a result of Gabber, they all have dualizing complexes. In particular, $\omega_X$ still exists and makes perfect sense.

We still have that $\mathcal{H}om_X(F_* O_X, \omega_X) \cong F_* \omega_X$ locally, but it is an open question whether this is a global isomorphism. It is possible to deduce that there exists a line bundle $L$ such that $$\mathcal{H}om_X(F_* O_X, \omega_X) \cong F_* (\omega_X \otimes L).$$ For some discussion of this, see THIS QUESTION OF MINE Regardless, in some applications, you can keep track of this line bundle. In particular, Frobenius splittings correspond to sections of some fixed twist of $\omega_X^{(1-p)}$.

The general scheme case:

The fact that $F_* O_X$ now need not be a finite $O_X$-module seems to make things much harder. Indeed, in this case, requiring a Frobenius splitting doesn't even seem to be the right thing to do.

However, in the local setting, commutative algebra has a replacement.

Definition (Hochster-Roberts): A local ring $R$ is called $F$-pure if for any module $M$, the map: $$M = M \otimes R \to M \otimes F_* R$$ is injective. This is a weaker condition than being $F$-split, and I think Mel Hochster told me an example at some point, but at this point I forget. Regardless,we have the following.

Theorem: An $F$-finite local ring is $F$-pure if and only if it is $F$-split.

A good reference for this is the paper of R. Fedder, $F$-purity and rationality singularity

It seems that this condition is much better behaved in a number of ways. If you search mathscinet for $F$-pure ring, you'll find many papers on this topic. I should point out that there is even a replacement for an ``honest'' Frobenius splitting in this context.

Set $E$ to be the injective hull of $R/\mathfrak{m}$. It turns out that $R$ is $F$-pure if and only if there exists an injective map $$E \to F_* E$$ (this is a recent result of Rodney Sharp).

Philosophical statement: Maps $E \to F_* E$ should be viewed as replacements for maps $F_* R \to R$ for non-$F$-finite rings.

If $R$ is $F$-finite, such maps are Grothendieck-local-dual to maps $F_* R \to R$ (at least for $R$ complete), and injective ones dualize to surjective ones (which can easily be tweaked into splittings). I think that even in the general context, such maps should still correspond to divisors, although I've never actually checked this.

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(You can say \begin{align} instead of $$, etc, and do the usual amsmath thing to get multiline equations, IIRC) –  Mariano Suárez-Alvarez Apr 3 '12 at 22:12
    
Thanks! That worked. –  Karl Schwede Apr 3 '12 at 22:21
    
Wow, this is a fantastic answer. Thanks! –  Chuck Hague Apr 4 '12 at 17:13

I thought about this a bit on the context of some conversations I was having with Allen Knutson, where I wanted to be able to localize to generic points of Frobenius split subvarieties. I came to the following conclusions:

(1) The relationship between near splittings and sections of $\omega_X^{-p+1}$ goes through the Cartier operator, which only works over perfect fields. To be concrete, suppose that $a \in k$ is not a $p$-th power. Consider the form $a (dx)^{-p+1}$ on $\mathbb{A}^1$. What near splitting would you like to associate it to?

If you work over perfect fields I could not find any problems.

(2) As I mentioned, I wanted to be able to localize at generic points of subvarieties, which will not be perfect. In this case, I had the following idea: Let $S$ be a ring of characteristic $p$ equipped with a splitting $\phi: S \to S$. I think the correct thing to study is varieties over $S$ equipped with splittings which extend $\phi$. In this case, I think we can then pass to any localization of $S$, quotient by any split ideal of $S$ and do other things we'd like to do. If $S$ is a perfect field then it has a unique splitting, given by $p$-th root, so this includes the case of my previous bullet point.

I didn't write anything about this because I didn't find anything interesting to do with it. The answer may just be that no one has had a reason to write this up. Or, of course, there could be some major issue I missed.

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Chuck, you may remember that when I asked Kumar why he hadn't written [Brion-Kumar] over perfect fields he said "My knowledge of perfect fields is very imperfect." –  Allen Knutson Apr 3 '12 at 20:27
    
This makes a lot of sense. And indeed I'm thinking mostly of Brion-Kumar when I ask this question; since the majority of the book uses the relationship between near-splittings and sections of $\omega_X^{-p+1}$ it makes sense that he would make assumptions on the base field from the beginning. @Allen -- Ah, I'd forgotten about that! –  Chuck Hague Apr 3 '12 at 21:02
    
Chuck, the relationship between near-splittings and such sections just needs the field to be $F$-finite (in other words, $k$ is a finite $k^p$ vector space). –  Karl Schwede Apr 3 '12 at 22:11
    
David, I would agree, the choice of of the Frobenius splitting of $F_* k \to k$ is basically the same as the choice of the isomorphism I mention below when I say: $$\Hom_X(F_* O_X, \omega_X) \cong F_* \omega_X$$ –  Karl Schwede Apr 3 '12 at 22:23

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